91Ó°ÊÓ

Two different testsAandBare to be given to a student chosen at random from a certain population. Suppose also that the mean score on testAis 85, and the standard deviation is 10; the mean score on testBis 90,and the standard deviation is 16; the scores on the two tests have a bivariate normal distribution, and the correlation of the two scores is 0.8.

Let A denote the first test scores, and let B denote second test scores.

Then,

\(\begin{array}{*{20}{l}}{{\mu _A} = 85\;}\\{{\sigma _A} = 10}\\{{\mu _B} = 90}\\{{\sigma _B} = 16}\\{p = 0.8}\end{array}\)

\(\)\(\)

For a\(BVN{\rm{ }}\left( {85,10,90,16,0.8} \right)\), the probability that her score on test A will be higher than her score on test B

\(P\left( {A > B} \right) = P\left( {A - B > 0} \right)\)

Let,\(C = A - B\)be a new variable.

The expectation of C is:

\(\begin{aligned}{}E\left( C \right) &= E\left( A \right) - E\left( B \right)\\ &= {\mu _A} - {\mu _B}\\ &= 85 - 90\\ &= - 5\end{aligned}\)

The standard deviation of C is:

\(\begin{aligned}{}\sqrt {Var\left( C \right)} &= \sqrt {Var\left( A \right) + Var\left( B \right) - 2 \times p \times sd\left( A \right) \times sd\left( B \right)} \\ &= \sqrt {{\sigma _A}^2 + {\sigma _B}^2 - 2 \times p \times {\sigma _A} \times {\sigma _B}} \\ &= \sqrt {100 + 256 - 2 \times 0.8 \times 10 \times 16} \\ &= \sqrt {100} \\ &= 10\end{aligned}\)

Therefore, C follows normal distribution, that is, \(C \sim N\left( { - 5,10} \right)\).

\(\begin{aligned}{}P\left( {C > 0} \right) &= P\left( {Z > \frac{{0 - \left( { - 5} \right)}}{{10}}} \right)\\& = P\left( {Z > 0.5} \right)\\ &= 1 - P\left( {Z \le 0.5} \right)\\ &= 1 - 0.6915\\& = 0.3085\end{aligned}\)

Therefore, the answer is 0.3085.